[LeetCode]38. Length of Last Word最后单词长度

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

For example, 
Given s = "Hello World",
return 5.

解法1：对输入字符串从后往前扫描，先去掉末尾空格得到输入字符串的“真正”end，然后找到第一个空格位置i，两个index之间即为最后一个单词，返回其长度即可。

class Solution {
public:
    int lengthOfLastWord(string s) {
        int n = s.size(), end = n - 1;
        int i = n - 1, j = n - 2;
        if (n == 0) return 0;
        for (; i >= 0; --i)
        {
            if (i == n - 1 && s[i] == ' ')
            {
                while (s[j] == s[j + 1]) --j;
                if (j < 0) return 0;
                i -= end - j;
                end = j;
            }
            if (s[i] == ' ') return end - i;
        }
        return end - i;
    }
};

解法2：另一个思路是先对输入字符串预处理，去掉开头和结尾多余的空格，然后从前往后扫描，遇到空格将计数器置零，否则计数器自增。这样计数器记录的自然就是最后一个单词的长度。

class Solution {
public:
    int lengthOfLastWord(string s) {
        int n = s.size(), count = 0;
        int left = 0, right = n - 1;
        while (left < n && s[left] == ' ') ++left;
        while (right >= 0 && s[right] == ' ') --right;
        for (int i = left; i <= right; ++i)
        {
            if (s[i] != ' ') ++count;
            else count = 0;
        }
        return count;
    }
};